# Proving the Nested Interval Property of the Reals

Using the sup(A) prove the Nested Interval Property of $\mathbb{R}$ We were given a sort of hint on this.

Let I$_{1}$ = [a$_{1}$,b$_{1}$], I$_{2}$ = [a$_{2}$,b$_{2}$],... be a sequence of nested closed (bounded) intervals. ($I_{1}\supset I_{2}\supset ...$) Then there exists at least one point common to every interal $\cap_{i=1}^{\inf}I_{i}\neq\oslash$

The $a_n$ are increasing. Show that if $c=\sup_n a_n$, then $a_n\leq c\leq b_n$ for every $n$.